Problem Statement: The sum of digit of two digit number is 9. Also 9 times this no is twice the no. obtained by reversing the order of the digits. Find the no. Explain in details.

Understanding the problem: We are given a two-digit number whose digits add up to 9. We need to find the number such that 9 times the number is twice the number obtained by reversing the order of the digits.

Solution:

Step 1: Let the two-digit number be represented as xy, where x and y are digits.

Step 2: As per the problem statement, the sum of digits is 9. Thus, x + y = 9.

Step 3: The number obtained by reversing the order of the digits is yx. Thus, the number obtained by reversing the order of the digits is 10y + x.

Step 4: 9 times the original number is 9xy.

Step 5: As per the problem statement, 9xy = 2(10y + x).

Step 6: Simplifying the equation, we get 9xy = 20y + 2x.

Step 7: Dividing both sides by 2, we get 4.5xy = 10y + x.

Step 8: Rearranging, we get 9xy = 20y + 2x.

Step 9: Substituting x + y = 9 in the above equation, we get 9y(9-y) = 20y + 18 - 2y.

Step 10: Simplifying, we get 9y^2 - 22y + 18 = 0.

Step 11: Factoring, we get (3y - 2)(3y - 9) = 0.

Step 12: Thus, y = 2/3 or y = 3.

Step 13: Since y is a digit, y can only be 3.

Step 14: Substituting y = 3 in x + y = 9, we get x = 6.

Step 15: Thus, the two-digit number is 63.

Verification: The sum of digits of 63 is 6 + 3 = 9. 9 times 63 is 567. The number obtained by reversing the order of the digits is 36. Twice of 36 is 72, which is equal to 567/9. Thus, the solution is correct.

Conclusion: The two-digit number is 63.